Temat: Dodawanie i odejmowanie ułamków o różnych mianownikach
Rozwiązanie i wyjaśnienie poniżej ;)
1)
[tex]3\dfrac{1}{2}+2\dfrac{3}{4}+1\dfrac{1}{3}=3\dfrac{1\cdot6}{2\cdot6}+2\dfrac{3\cdot3}{4\cdot3}+1\dfrac{1\cdot4}{3\cdot4}=3\dfrac{6}{12}+2\dfrac{9}{12}+1\dfrac{4}{12}=6\dfrac{19}{12}=\boxed{\boxed{7\frac{7}{12}}}\\[/tex]
2)
[tex]2\dfrac{1}{20}-\dfrac{3}{4}=2\dfrac{1}{20}-\dfrac{3\cdot5}{4\cdot5}=2\dfrac{1}{20}-\dfrac{15}{20}=1\dfrac{21}{20}-\dfrac{15}{20}=1\dfrac{6}{20}=\boxed{\boxed{1\dfrac{3}{10}}}\\[/tex]
3)
[tex]1\dfrac{7}{18}-2\dfrac{2}{9}=1\dfrac{7}{18}-2\dfrac{2\cdot2}{9\cdot2}=1\dfrac{7}{18}-2\dfrac{4}{18}=\dfrac{25}{18}-\dfrac{40}{18}=-\dfrac{15}{18}=\boxed{\boxed{-\dfrac{5}{6}}}\\[/tex]
4)
[tex]2\dfrac{1}{7}-5\dfrac{2}{3}=-(5\dfrac{2\cdot7}{3\cdot7}-2\dfrac{1\cdot3}{7\cdot3})=-(5\dfrac{14}{21}-2\dfrac{3}{21})=\boxed{\boxed{-3\dfrac{11}{21}}}\\[/tex]
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Temat: Dodawanie i odejmowanie ułamków o różnych mianownikach
Rozwiązanie i wyjaśnienie poniżej ;)
1)
[tex]3\dfrac{1}{2}+2\dfrac{3}{4}+1\dfrac{1}{3}=3\dfrac{1\cdot6}{2\cdot6}+2\dfrac{3\cdot3}{4\cdot3}+1\dfrac{1\cdot4}{3\cdot4}=3\dfrac{6}{12}+2\dfrac{9}{12}+1\dfrac{4}{12}=6\dfrac{19}{12}=\boxed{\boxed{7\frac{7}{12}}}\\[/tex]
2)
[tex]2\dfrac{1}{20}-\dfrac{3}{4}=2\dfrac{1}{20}-\dfrac{3\cdot5}{4\cdot5}=2\dfrac{1}{20}-\dfrac{15}{20}=1\dfrac{21}{20}-\dfrac{15}{20}=1\dfrac{6}{20}=\boxed{\boxed{1\dfrac{3}{10}}}\\[/tex]
3)
[tex]1\dfrac{7}{18}-2\dfrac{2}{9}=1\dfrac{7}{18}-2\dfrac{2\cdot2}{9\cdot2}=1\dfrac{7}{18}-2\dfrac{4}{18}=\dfrac{25}{18}-\dfrac{40}{18}=-\dfrac{15}{18}=\boxed{\boxed{-\dfrac{5}{6}}}\\[/tex]
4)
[tex]2\dfrac{1}{7}-5\dfrac{2}{3}=-(5\dfrac{2\cdot7}{3\cdot7}-2\dfrac{1\cdot3}{7\cdot3})=-(5\dfrac{14}{21}-2\dfrac{3}{21})=\boxed{\boxed{-3\dfrac{11}{21}}}\\[/tex]